To prepare 2L of a 5M solution, you should put in 4.6575grams of KMnO4. It is important to make sure that you add them in that order. K should be added first, then Mn.
To make a 3.7% EDTA solution, you would add 3.7 grams of EDTA to 100 mL of solution.
This depends on the volume and concentration of this solution.
This solution contain 26,3 g NaOH.
There would be 946 mg (2 mg/ml x 473 ml) of active ingredient in 473 ml of the solution. To convert milligrams to grams, divide by 1000, so 946 mg is equal to 0.946 grams in 473 ml of the solution.
(158 g = 1 mole) --- molar mass of potassium permanganate. You also need to specify the volume to be made. For 1 liter just add 15.8 g in a volumetric flask to make 1000 ml (1 liter) of solution.
To prepare 2L of a 5M solution, you should put in 4.6575grams of KMnO4. It is important to make sure that you add them in that order. K should be added first, then Mn.
The compound potassium permanganate has chemical formula KMnO4 Molecular mass of KMnO4 = 39.1 + 54.9 + 4(16.0) = 158.0 Mass of KMnO4 = amount of KMnO4 x Molecular mass of KMnO4 = 2.55 x 158.0 = 403g
Molarity = moles of solute/Liters of solution ( 75.0 ml = 0.075 Liters ) Algebraically manipulate, moles of solute = Liters of solution * Molarity Moles KMnO4 = (0.075 Liters)(0.0950 M) = 7.13 X 10 -3 moles KMnO4 ------------------------------------
2 TABLETS
To calculate the grams of potassium permanganate in 2.20 moles, you would need to know the molar mass of potassium permanganate. The molar mass of potassium permanganate (KMnO4) is about 158.034 g/mol. So, 2.20 moles of KMnO4 is equal to 2.20 moles x 158.034 g/mol = 347.67 grams of potassium permanganate.
the charge in KMno4 is zero . it has no unpaired electrons but it is still dark in colored due to charge transfer in KMnO4. from Ayushi Sharma
40.8 grams
438 grams.
To calculate the grams of phosphate in a solution, you first need to determine the molarity of the solution. Once you know the molarity, you can use the molecular weight of phosphate to determine the grams present in the solution. Can you provide the concentration or volume of the K2HPO4 solution?
To prepare a 3% solution of sulfosalicylic acid, you would need 30 grams of sulfosalicylic acid for every 1 liter of solution.
The answer is 8 g NaCl.